How much heat is required to convert 10g of ice at -20°C into steam at 100°C is(specific heat of ice=0.5cal/g-°C?

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Apr 12, 2017

$7300$ calories

Explanation:

When $10 g$ of ice at -20°C s being converted into steam at 100°C, there are four stages.

1. Ice at -20°C to ice at 0°C - here it continues to be in the same state i.e. ice and hence heat required is $\text{mass"xx"specific heat"xx"change in temperature}$ Specific heat for ice is $0.5$ cal/g-°C.
2. Then from ice at 0°C to water at 0°C and heat required is $\text{mass"xx"latent heat}$, this latent heat required is for conversion of each unit mass of substance. From ice to water it is $80 \text{cal per gram}$
3. Water at 0°C to water at 100°C - here it continues to be in the same state i.e. water and hence heat required is $\text{mass"xx"specific heat"xx"change in temperature}$ Specific heat for water is $1$ cal/g-°C.
4. Then from water at 100°C to steam at 100°C and heat required is $\text{mass"xx"latent heat}$, this latent heat required is for conversion of each unit mass of substance. From water to steam, it is $540 \text{cal per gram}$

Hence, to find heat is required to convert $10 g$ of ice at -20°C into steam at 100°C,

first calculate heat required to convert $10 g$ of ice at -20°C to ice at 0°C. As specific heat of ice is $0.5 \text{cal/g°C}$, this is

$10 \times 20 \times 0.5 = 100$ cal.

heat required to convert $10 g$ of ice to $10 g$ of water at 0°C is

$10 \times 80 = 800$ cal - as latent heat is $80 \text{cal/g}$

heat required to convert $10 g$ of water at 0°C to $10 g$ of water at 100°C is

$10 \times 100 x 1 = 1000$ cal

heat required to convert $10 g$ of water at 1000°C to $10 g$ of steam at 100°C is

$10 \times 540 = 5400$ cal - as latent heat is $540 \text{cal/g}$

Hence, total heat required is $100 + 800 + 1000 + 5400 = 7300$ calories

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